Problem: Kana was just an ordinary high school girl before a talent scout discovered her. Then, she became an idol. But different from the stereotype, she is also a gameholic. 

One day Kana gets interested in a new adventure game called Dragon Quest. In this game, her quest is to beat a dragon.

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The dragon has a hit point of x initially. When its hit point goes to 0 or under 0, it will be defeated. In order to defeat the dragon, Kana can cast the two following types of spells. 

  * Void Absorption

Assume that the dragon's current hit point is h, after casting this spell its hit point will become \left⌊ h/2 \right⌋ + 10. Here \left⌊ h/2 \right⌋ denotes h divided by two, rounded down.

  * Lightning Strike

This spell will decrease the dragon's hit point by 10. Assume that the dragon's current hit point is h, after casting this spell its hit point will be lowered to h-10.




Due to some reasons Kana can only cast no more than n Void Absorptions and m Lightning Strikes. She can cast the spells in any order and doesn't have to cast all the spells. Kana isn't good at math, so you are going to help her to find out whether it is possible to defeat the dragon.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The next t lines describe test cases. For each test case the only line contains three integers x, n, m (1≤ x ≤ 10^5, 0≤ n,m≤30) — the dragon's intitial hit point, the maximum number of Void Absorptions and Lightning Strikes Kana can cast respectively.

Output

If it is possible to defeat the dragon, print "YES" (without quotes). Otherwise, print "NO" (without quotes).

You can print each letter in any case (upper or lower).

Example

Input


7
100 3 4
189 3 4
64 2 3
63 2 3
30 27 7
10 9 1
69117 21 2


Output


YES
NO
NO
YES
YES
YES
YES

Note

One possible casting sequence of the first test case is shown below:

  * Void Absorption \left⌊ 100/2 \right⌋ + 10=60.
  * Lightning Strike 60-10=50.
  * Void Absorption \left⌊ 50/2 \right⌋ + 10=35.
  * Void Absorption \left⌊ 35/2 \right⌋ + 10=27.
  * Lightning Strike 27-10=17.
  * Lightning Strike 17-10=7.
  * Lightning Strike 7-10=-3.

Solution (in PYTHON):

Solution:
from __future__ import division, print_function
# import threading
# threading.stack_size(2**27)
# import sys
# sys.setrecursionlimit(10**7)
from sys import stdin, stdout
import bisect            #c++ upperbound
import math
import heapq
i_m=9223372036854775807
def modinv(n,p):
    return pow(n,p-2,p)
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
import math  
def GCD(x, y): 
    x=abs(x)
    y=abs(y)
    if(min(x,y)==0):
        return max(x,y)
    while(y): 
        x, y = y, x % y 
    return x 
def Divisors(n) : 
    l = []  
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                l.append(n//i)
    return l
 
q=[]       
def dfs(n,d,v,c):
    global q
    v[n]=1
    x=d[n]
    q.append(n)
    j=c
    for i in x:
        if i not in v:
            f=dfs(i,d,v,c+1)
            j=max(j,f)
            # print(f)
    return j
def SieveOfEratosthenes(n): 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    f=[]
    for p in range(2, n): 
        if prime[p]: 
            f.append(p)
    return f
  
"""*******************************************************"""
def main():
    t=inin()
    for _ in range(t):
        n,m,k=cin()
        for i in range(m):
            f=n//2+10
            if(f<n):
                n=f
            else:
                break
        k=k*10
        if(n<=k):
            print("YES")
        else:
            print("NO")

 
######## Python 2 and 3 footer by Pajenegod and c1729
 
# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.
 
# So on cf, use PyPy2 for best string performance.
 
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
 
import os, sys
from io import IOBase, BytesIO
 
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
 
# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'R' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'R' [0]:
        A.append(sign*numb)
    return A
 
# threading.Thread(target=main).start()
if __name__== "__main__":
  main()